How Does Quantum Mechanics Describe Momentum in Three Dimensions?

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SUMMARY

The discussion focuses on the projection of the wave function in three-dimensional momentum space using Fourier transforms. The key equation presented is the three-dimensional inner product, expressed as &langle \vec{p}|\psi \rangle = \int d^{3}\vec{r} \langle \vec{p}|\vec{r} \rangle \langle \vec{r}|\psi \rangle. The participants clarify the significance of terms like &langle x|p_{y}\rangle and derive the relationship &langle \vec{p}|\vec{r} \rangle = \langle p_x|x \rangle \langle p_y|y \rangle \langle p_z|z \rangle through the generalization of the Fourier Transform in three dimensions. The discussion emphasizes the importance of understanding the linearity of inner products and the role of the position and momentum operators in this context.

PREREQUISITES
  • Understanding of Quantum Mechanics, specifically wave functions and their projections.
  • Familiarity with Fourier Transforms in multiple dimensions.
  • Knowledge of Dirac notation and inner product properties.
  • Basic concepts of linear algebra as applied to quantum states.
NEXT STEPS
  • Study the derivation of the Fourier Transform in three dimensions.
  • Explore the implications of the commutation relations [x_{l},p_{m}] = i\hbar \delta_{l,m}.
  • Learn about the expansion of position and momentum operators in terms of their components.
  • Investigate the application of linearity in quantum mechanics and its effects on inner products.
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Students and professionals in quantum mechanics, physicists focusing on wave-particle duality, and researchers interested in the mathematical foundations of quantum theory.

varunag
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In our Quantum mechanics course we were told that the wave function can be projected in the momentum space by taking the Fourier transform of the projection of the wave function in the coordinate space. This was shown in one-dimension as:
\langle p|\psi \rangle = \int_{-\infty}^{\infty} \langle p|x \rangle \langle x|\psi \rangle dx

But, how do we write if we hve to write in three dimensions?
i.e.
\langle \vec{p}|\psi \rangle = \int_{-\infty}^{\infty} \langle \vec{p}|\vec{r} \rangle \langle \vec{r}|\psi \rangle dx
Here I'm confused as to what is \langle \vec{p}|\vec{r} \rangle ?
If I write \vec{p} = p_x \hat{i} + p_y \hat{j} + p_z \hat{k}
and \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}
then should the terms like, \langle p_y|x \rangle crop up in the final result?
 
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In 3 dimensions, the position operator \hat{x} is conventionally denoted by \hat{r} (since the position vector in 3-space is denoted by \vec{r}). Having said that you can still use \hat{x} with the understanding that \hat{x} has three components and that the integral you have written down has to be carried out over each of the 3 spatial coordinates.

Using \hat{r}, I will write this as

\langle p|\psi\rangle = \int d^{3}\vec{r} \langle p|r\rangle \langle r|\psi\rangle

Now, I can use

\langle r|p \rangle = \frac{1}{(2\pi \hbar)^{3/2}}\exp\left(\frac{i}{\hbar}{\vec{p}\bullet \vec{r}}\right)

But I think what you wanted to ask was the significance of the quantity \langle x|p_{y}\rangle.

EDIT: To do it rigorously, you can use linearity of the inner product:

\langle \alpha_{1} + \alpha_{2} | \beta_{1} \rangle = \langle \alpha_{1} | \beta_{1} \rangle + \langle \alpha_{2} | \beta_{1}\rangle

with suitable number of terms, using the expansion of the position and momentum operators in terms of their x, y, z components and then using the expressions for \langle x|p_{x}\rangle, and so on. You will get the cross term \langle x|p_{y}\rangle in such an expression.
 
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The result you have given is known, but I don't know how the result came. i.e. how do we write:
\langle \vec{p}|\vec{r} \rangle = \langle p_x|x \rangle \langle p_y|y \rangle \langle p_z|z \rangle?
 
varunag said:
The result you have given is known, but I don't know how the result came. i.e. how do we write:
\langle \vec{p}|\vec{r} \rangle = \langle p_x|x \rangle \langle p_y|y \rangle \langle p_z|z \rangle?

This follows from the generalization of the Fourier Transform in 3 dimensions. Specifically,

\psi(\vec{r}) = \frac{1}{(2\pi\hbar)^{3/2}} \int d^{3}\vec{p} \exp\left(-\frac{i}{\hbar}\vec{p}\bullet\vec{r}\right)\phi(\vec{p})

\phi(\vec{p}) = \frac{1}{(2\pi\hbar)^{3/2}} \int d^{3}\vec{r} \exp\left(\frac{i}{\hbar}\vec{p}\bullet\vec{r}\right)\psi(\vec{r})

You can write this in Dirac notation easily.

Note: The result is consistent with the fact that [x_{l},p_{m}] = i\hbar \delta_{l,m}.

EDIT: This can be derived by inserting complete sets of states, i.e. writing \langle \vec{p}|\vec{r}\rangle as an integral (multiple integral) over basis states corresponding to the different x, y, z components of p and r.
 

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